Question 140321
Start with the given equations


{{{x^2+y^2=26}}}
{{{y=-x+8}}}





{{{x^2+y^2=26}}} Start with the first equation


{{{x^2+(-x+8)^2=26}}} Plug in {{{y=-x+8}}}



{{{x^2+x^2-16x+64=26}}} Foil



{{{x^2+x^2-16x+64-26=0}}} Subtract 26 from both sides



{{{2x^2-16x+38=0}}} Combine like terms



Let's use the quadratic formula to solve for x:



Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general solution using the quadratic equation is:


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}




So lets solve {{{2*x^2-16*x+38=0}}} ( notice {{{a=2}}}, {{{b=-16}}}, and {{{c=38}}})





{{{x = (--16 +- sqrt( (-16)^2-4*2*38 ))/(2*2)}}} Plug in a=2, b=-16, and c=38




{{{x = (16 +- sqrt( (-16)^2-4*2*38 ))/(2*2)}}} Negate -16 to get 16




{{{x = (16 +- sqrt( 256-4*2*38 ))/(2*2)}}} Square -16 to get 256  (note: remember when you square -16, you must square the negative as well. This is because {{{(-16)^2=-16*-16=256}}}.)




{{{x = (16 +- sqrt( 256+-304 ))/(2*2)}}} Multiply {{{-4*38*2}}} to get {{{-304}}}




{{{x = (16 +- sqrt( -48 ))/(2*2)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (16 +- 4*i*sqrt(3))/(2*2)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{x = (16 +- 4*i*sqrt(3))/(4)}}} Multiply 2 and 2 to get 4




After simplifying, the quadratic has roots of


{{{x=4 + sqrt(3)i}}} or {{{x=4 - sqrt(3)i}}}




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Answer:


Since we get a complex solution for the equation {{{2x^2-16x+38=0}}}, this means that the equations {{{x^2+y^2=26}}} and {{{y=-x+8}}} do not intersect. So there are no solutions